Se [tex]a[/tex] e [tex]b[/tex] são números reais, então [tex]2ab\leq a^2+b^2;[/tex] se [tex]a[/tex] e [tex]b[/tex] forem não negativos, então [tex]\sqrt{ab} \leq \dfrac{1}{2} (a+b)[/tex] e ocorre a igualdade se, e somente se, [tex]a=b[/tex].
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[tex]\displaystyle \sf \text{Sejam a e b}\in\mathbb{R}.\ \ \text{Fa\c camos} :\\\\ (a-b)^2\geq 0 \\\\ a^2-2ab+b^2\geq 0 \\\\ \Large\boxed{\sf \ a^2+b^2 \geq 2ab \ }\checkmark \text{C.Q.D}[/tex]
se a = b, temos :
[tex]\sf a^2+b^2\geq 2ab \\\\ a^2+a^2\geq 2a^2 \\\\ 2a^2 \geq 2a^2 \\\\ \boxed{\sf \ a^2=a^2 \ }\checkmark[/tex]
[tex]\displaystyle \sf \text{Sejam a e b }\in\mathbb{R}_{+}\ . \ \text{Fa\c camos} : \\\\ \left(\sqrt{a}-\sqrt{b}\right)^2 \geq 0 \\\\ a-2\sqrt{ab}+b\geq 0 \\\\ a+b\geq 2\sqrt{ab} \\\\ \Large\boxed{\sf \ \frac{a+b}{2}\geq \sqrt{ab}\ }\checkmark \text{C.Q.D }[/tex]
se a = b, temos :
[tex]\displaystyle \sf \frac{a+b}{2}\geq \sqrt{ab} \\\\\ a = b : \\\\ \frac{a+a}{2}\geq \sqrt{a\cdot a}\geq \sqrt{a^2} \\\\\\ a\geq a \\\\\ \Large\boxed{\sf \ a = a\ }\checkmark[/tex]